A beautiful coin game

[tuanl]

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process. Calculate P(X>Y).

 

[Adrián Facchi]

Here is my guess.
To have \(X=n\) we need a first set of \(n-2\) tails and \(2\) heads, so the probability is
\(P(X=n)=\frac{1}{2^n}\ ,\qquad n\geqslant 2 .\)
And analogously for\(Y\),we need \(n-3\) tails and \(3\) heads
\(P(Y=n)=\frac{1}{2^n}\ , \qquad n\geqslant 3 .\)
However we have always \(X\geqslant 2\) and \(Y\geqslant 3\) so actually we can say that
\(P(X \geqslant Y|X\geqslant 3) = \frac{1}{2} \qquad \text{and}\qquad P(X\geqslant Y|X=2)=0 .\)
Therefore
\(P(X\geqslant Y) = P(X\geqslant 3)P(X\geqslant Y|X\geqslant 3) = \frac{3}{4}\times\frac{1}{2}= \frac{3}{8}\)
Note that this is a solution for \(P(X \geqslant Y)\), the strict inequallity probably needs a little bit more thinking.

 

[Analytic Bastard]

On Adrian's solution:
P(X>= 3)=P(X<3)=1-P(X=1)-P(X=2)=1-1/2-1/4=1-3/4=1/4 not the 3/4 computed.

As for P(X>Y|X>=3), it is 0 for X=3, (since we are taking X greater than Y and Y can only be at least 3) and 1/2 when X=4, but, for example, it is 3/4 for X=5.

I must go AFK now but I'll come back to the problem.

 

[Adrián Facchi]

On Adrian's solution:
P(X>= 3)=P(X<3)=1-P(X=1)-P(X=2)=1-1/2-1/4=1-3/4=1/4 not the 3/4 computed.

As for P(X>Y|X>=3), it is 0 for X=3, (since we are taking X greater than Y and Y can only be at least 3) and 1/2 when X=4, but, for example, it is 3/4 for X=5.

I must go AFK now but I'll come back to the problem.

Well you are right except that I assumed that the tosses that give the result are also counted as number of tosses so (P(X=1)=0) in your notation.

In relation with the other comment, you are right. Without realizing I actually computed the probability of (P(X\geqslant Y)) not the strict inequallity. I will edit my solution.

 

[Analytic Bastard]

Well you are right except that I assumed that the tosses that give the result are also counted as number of tosses so \(P(X=1)=0\) in your notation.

In relation with the other comment, you are right. Without realizing I actually computed the probability of \(P(X\geqslant Y)\) not the strict inequallity. I will edit my solution.

Adrian, you are right in the first part, it is indeed \(P(X\geq 3)=\frac{3}{4}\) since \(P(X=1)=0\), sorry about that.

However, I just see a total probability approach here.
\(P(X>Y)=P(\max(X,Y)=X)=\sum\limits_{x=3}^{\infty} P(\max(X,Y)=X|X=x)P(X=x)\)
Now, \(P(\max(X,Y)=X|X=x) = P(Y\leq x)\)

Therefore
\(P(\max(X,Y)=X|X=x)P(X=x)=P(Y\leq x)P(X=x)\)

So that
\(P(X>Y)=\sum\limits_{x=3}^{\infty} P(Y\leq x)P(X=x)\)

Which would certainly yield the probability you look for. Counting, however, is not my streght.

Cmon people, keep it up.

 

[Analytic Bastard]

Kids,

I modelled the problem using Markov chains. I made a state space model consisting of 10 states, including the starting one (no toss). Three absorbing states were added (HH win, HH - HHH tie and HHH win, where winning means reaching the goal first and therefore with a smaller number of tosses). The fact that HH wins means that the number X in the statement is less than Y. Therefore, applying the absortion probability equations for state HHH win gets the result. It is a fairly laborious exercise for my sparte time. If anybody is interested I can tell you my state space model under the promise to post the solution.

 

[Analytic Bastard]

Well I leave here the transition probability matrix

0.2500 0.2500 0.2500 0.2500 0 0 0 0 0 0
0.2500 0 0 0 0.5000 0.2500 0 0 0 0
0.2500 0 0 0.2500 0 0.2500 0.2500 0 0 0
0.2500 0 0.2500 0 0.5000 0 0 0 0 0
0 0 0 0 1.0000 0 0 0 0 0
0.2500 0 0 0 0.2500 0 0 0.2500 0.2500 0
0.2500 0 0 0.2500 0 0 0 0.5000 0 0
0 0 0 0 0 0 0 1.0000 0 0
0 0 0 0 0 0 0 0 1.0000 0
0.2500 0.2500 0.2500 0.2500 0 0 0 0 0 0

Row i contains the transition probability from i to j. The starting state is 10. HHH win is number 8. States 5 (HH win) and 9 (tie) are also absorbent.

 

[JinWJ]

Here's my shot,


Define \(X=X'+1, Y=Y'+2\)
Then \(X' \sim Geom(\frac{1}{4})\) and \(Y' \sim Geom(\frac{1}{8})\)
By a linear change of variable,
\(f_{X'} = (1-p)^{x'}p \)
\(f_{Y'} = (1-p)^{y'+1}p\)
\(P(X>Y) = P(X'>Y'+1) = \sum\limits_{x=1}^{\infty}P(X'=x)P(Y'<x-1) = \ldots = \frac{2}{11}\)